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MUSICAL ACOUSTICS Tutorial 2: Solution Guide

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MUSICAL ACOUSTICS Tutorial 2: Solution Guide MUSICAL ACOUSTICS Tutorial 2: Solution Guide 1. Speed of sound in air The speed of sound c in dry air is given approximately (in ms−1) by the formula c = 331.3 + 0.606T where T is the temperature in degrees Celsius. Calculate the frequency of a sinusoidal sound with a wavelength in air λ = 1m when the airtemperature is (a) 0 ◦C; (b) 22 ◦C. What is the pitch difference between these sounds? (a) For T = 0 ◦C, the speed of sound in air is c = 331.3ms−1. So the frequency is f (0) = c/λ = 331.3/1 = 331.3Hz. (b) For T = 22 ◦C, the speed of sound in air is c = 331.3 + 0.606 × 22 = 344.9ms−1. So the frequency is f (22) = c/λ = 344.6/1 = 344.6ms−1. The frequency ratio is f (22)/ f (0) = 344.6/331.3 = 1.0401. So the pitch interval in cents is 1200 f (22) ∆P = log = 3986log(1.0401) = 68cents. log(2) f (0) 2. Resonance of ear canal The average length of the adult human ear canal is around 25mm. If the ear canal is modelled as a cylindrical tube closed at the eardrum and open at the outer end, what is its lowest resonance frequency? What is the approximate pitch of this resonance? (Note: human body temperature ≃ 37 ◦C). For the lowest resonance of a cylindrical tube of length L closed at one end and open at the other, λ = 4L. So if L = 25mm, λ = 100mm = 0.1m. The frequency of this resonance is given by f = c/λ. In the ear canal, the air has a temperature of approximately 37 ◦C. The speed of sound at this temperature is given by the formula in Question 1 as c = 331.3 + 0.606 × 37 = 353.7ms−1. Thus the frequency is f = 353.7/0.1 = 3537Hz. You could look this up in a table of equal temperament frequencies (e.g. Musician’s Guide to Acoustics p.178, or http://en.wikipedia.org/wiki/Piano key frequencies). You will find the nearest piano note is A7. More accurately, you could use the pitch difference formula to compare the pitch with one whose frequency you know, such as A4 (frequency 440 Hz): 1200 3537 ∆P = log = 3986log(8.039) = 3608cents. log(2) 440 Since there are 1200 cents in one octave, 3608 cents is three octaves plus 8 cents, so the calculated resonance frequency corresponds to a pitch 8 cents higher than A7. 1 3. Equally tempered intervals How many cents are there in the following equally tempered intervals: (a) octave; (b) semitone; (c) major third; (d) perfect fifth? An equally tempered interval always contains an integer multiple of 100 cents, the equally tempered semitone. ET interval: octave semitone major third perfect fifth No. of semitones: 12 1 4 7 No. of cents: 1200 100 400 700 4. Just intonation intervals Just intonation intervals are those found between different members of a harmonic series, such as that shown in Figure 1. Figure 1: Intervals between members of a harmonic series. Harmonic number shown above the notes; intervals (in cents) shown below. How many cents are there in the following just intonation intervals: (a) octave; (b) semitone; (c) major third; (d) perfect fifth? (Hint: find these intervals in Figure 1.) JI interval: octave semitone major third perfect fifth Between harmonics: 1and 2 15 and 16 4 and 5 2 and 3 No. of cents: 1200 112 386 702 You can check the intervals by using the frequency ratios in the formula: for example, a major third has a frequency ratio of 5/4 = 1.25, so the pitch interval in cents is ∆P = 3986 × log(1.25) = 386 cents. 5. Critical bands For frequencies above a few hundred hertz, the critical bandwidth of the human ear is approximately one third of an octave. What is the pitch interval corresponding to 1/3 octave? How high up the harmonic series in Figure 1 do you need to go to find adjacent harmonics lying within one critical band? What is the perceptual significance of this? An octave contains 12 semitones, so a 1/3 octave band corresponds to an interval of 4 semitones, or an ET major third. Above the 4th harmonic, adjacent pairs of harmonics lie inside one critical band, since they are less than a major third apart. This means that the timbre of sounds with strong harmonics above the 4th will be affected by the roughness or beating between adjacent harmonics. 2 6. Decibels and Sound Pressure Level A trumpet player creates a Sound Pressure Level of 70 dB in a concert hall. To what musical dynamic marking does this SPL correspond? How would the SPL value and the dynamic level change if (a) a second trumpeter joined in with the same sound power as the first; (b) ten trumpeters played in unison, each with the same sound power as the first? From the table of dynamic levels in the lecture notes on Anatomy of a musical note we can estimate that a 70dB trumpet note will sound mf. If a second trumpeter joins in, the sound power and therefore the sound intensity will double. The decibel change is given by the formula change = 10logI2/I1. Here I2/I1 = 2, so the decibel increase is 10 ×0.30 = 3dB: theSPL willgo up from 70dB to 73dB. This is about a third of one dynamic step. If 10 trumpeters play simultaneously, the intensity will go up by a factor of 10. Substituting I2/I1 = 10 in the formula gives the decibel increase as 10 × 1 = 10dB. The SPL will go up to 80dB, corresponding to an increase in dynamic level from mf to f. 7. Inverse square law In the open air, the intensity generated by an isotropic sound source is inversely proportional to the square of the distance from the source. What does “isotropic” mean? If you stand 3m from a bagpipe player on a flat moor and measure a Sound Pressure Level of 90dB, what SPL will you measure at a distance of (a) 30m; (b) 300m? An isotropic sound source is one which radiates sound with equal intensity in all directions. Assuming that the bagpipe player can be treated as an isotropic source (which could be a subject for discussion), the intensity will fall off with the square of the distance. (a) Going from 3m to 30m increases the distance by a factor 10, and therefore decreases the inten- −2 −2 sity by a factor 10 . Using the decibel formula in Question 6 with I2/I1 = 10 gives − change = 10log(10 2) = 10 × (−2) = −20dB. The SPL will drop from 90dB to 70dB. (b) Going from 3m to 300m increases the distance by a factor 102, and therefore decreases the −4 −4 intensity by a factor 10 . Using the decibel formula in Question 6 with I2/I1 = 10 gives − change = 10log(10 4) = 10 × (−4) = −40dB. The SPL will drop from 90dB to 50dB. 3 8. Masking In the first bar of a musical score, a flute plays the note C4 with an SPL of 70dB and a trombone simultaneously plays the note C3 at an SPL of 80dB. In the next bar the flute keeps playing at the same level, but the trombone is replaced by a trumpet playing the note C5 at an SPL of 80dB. Explain why the flute will sound quieter in the first bar than in the second bar. Flute mf Trumpet in C f Trombone f Figure 2: Effect of masking by lower and higher instruments. In the musical example, shown in Figure 2, the flute is accompanied in each bar by another instru- ment playing at a higher dynamic level. This is a circumstance in which the flute sound can be masked by the other instrument: the flute can sound quieter than it would in the absence of the masking instrument (partial masking), or it can become completely inaudible (total masking). However, the frequency components of the flute which are masked most effectively are those above the frequencies generated by the masker. When the masker is the trombone, all the flute’s harmonics are above the trombone’s fundamental, so masking will be very strong. When the masker is the trumpet, the lowest masking frequency is an octave above the flute’s fundamental, so masking will be less effective. 4.
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